Question

In the given figure $AD=CD$ and $BD=AD$, then what is the measure of $\angle ABC$ ?

• A. ${40}^o$
• B. ${50}^o$
• C. ${60}^o$
• D. ${70}^o$

Answer 1

The correct option is A ${40}^o$

$\angle ACE+\angle ACD=180$
$130+\angle ACD=180$
$\angle ACD=50$
Since, AD = DC
thus, $\angle DAC=\angle ACD=50$
In $△ADC$
$\angle DAC+\angle ACD+\angle ADC=180$
$50+50+\angle ADC=180$
$\angle ADC={80}^{\circ }$
Now, $\angle ADC+\angle ADB=180$
$80+\angle ADB=180$
$\angle ADB=100$
Since, $BD=AD$, $\angle BAD=\angle DBA$ (angles of equal and opposite sides are equal)
In $△$ $BDA$,
$\angle BAD+\angle DBA+\angle ADB=180$
$2\angle ABD=180-100$
$\angle ABD=\angle ABC={40}^{\circ }$